Is this a complete test to show that a matrix is orthogonal?

I used to test orthogonality by using the definition MT = M-1, which means I always calculated the inverse of the matrices. However, isn't it true that if M is orthogonal, then MMT = I?

If we multiply both side by M-1, we get MT = M-1.

Can I use this to proof the orthogonality of a matrix M, instead of calculating it's (often tedious) inverse?
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 Quote by tamtam402 I used to test orthogonality by using the definition MT = M-1, which means I always calculated the inverse of the matrices. However, isn't it true that if M is orthogonal, then MMT = I? If we multiply both side by M-1, we get MT = M-1. Can I use this to proof the orthogonality of a matrix M, instead of calculating it's (often tedious) inverse?

Of course...it's exactly the same, right?!

DonAntonio

 Quote by DonAntonio Of course...it's exactly the same, right?! DonAntonio
This might be a dumb question, but would it be possible to have a matrix M where MMT = I, yet MTM ≠ I? I'm pretty sure that by definition, proving either of these 2 equalities is enough to know that M-1 = MT, but I wanted to make sure.

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Is this a complete test to show that a matrix is orthogonal?

 Quote by tamtam402 This might be a dumb question, but would it be possible to have a matrix M where MMT = I, yet MTM ≠ I? I'm pretty sure that by definition, proving either of these 2 equalities is enough to know that M-1 = MT, but I wanted to make sure.
That is actually a very good question. There is no reason why $MM^T=I$ should imply $M^TM=I$.
But this is actually quite a deep result in linear algebra. The result is that every left-invertible matrix is invertible. And likewise with right-invertible.
So if AB=I, then it holds that BA=I. This is a special result that only holds for matrices.

 Quote by micromass That is actually a very good question. There is no reason why $MM^T=I$ should imply $M^TM=I$. But this is actually quite a deep result in linear algebra. The result is that every left-invertible matrix is invertible. And likewise with right-invertible. So if AB=I, then it holds that BA=I. This is a special result that only holds for matrices.

Well, that holds in any group and in any monoid where every element has a right inverse, since then ( with $\,a'=\,$ right inverse of $\,a$ ):
$$a'a=a'a(a'a'')=a'(aa')a''=a'ea''=a'a''=e$$
so the right inverse is also the left one.

DonAntonio